JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
Given below are two statements:
Statement I : H₂Se is more acidic than H₂Te
Statement II : H₂Se has higher bond enthalpy for dissociation than H₂Te
In the light of the above statements, choose the correct answer from the options given below:
- (a) Both Statement I and Statement II are true
- (b) Statement I is true but Statement II is false
- (c) Both Statement I and Statement II are false
- (d) Statement I is false but Statement II is true
Answer and solution
Answer (d) Statement I is false but Statement II is true
Solution
Chemistry ma’am · 4 steps
Going down Group 16 the H–E bond gets weaker, and the weaker bond gives up its H⁺ more easily.
- Size of the atom. Te is below Se in Group 16, so it is a bigger atom. Its valence orbitals are larger and more diffuse, so they overlap poorly with the small 1s orbital of hydrogen.
- Bond enthalpy. Poorer overlap means a weaker bond. So the H–Se bond is stronger than the H–Te bond, and Statement II is true.
- Acidic strength. An acid works by releasing H⁺, which means breaking the H–E bond. The weaker H–Te bond breaks more easily, so H₂Te is the stronger acid. That makes Statement I false.
Hydride H–E bond Acidic strength H₂O strongest weakest H₂S weaker stronger H₂Se stronger than H–Te weaker than H₂Te H₂Te weakest strongest - So. Statement I is false and Statement II is true, which is option (d).
The idea to keep. For hydrides down a group, size wins over electronegativity: bond enthalpy falls, so acidic strength rises. H₂O < H₂S < H₂Se < H₂Te in acidity.
Answer from the NTA final answer key. Solution by PrepWiser.